*Determine whether the following assertions are true or false. if true, prove the result, and if false, give a counter example

2. If p is a prime and p/(a^2+b^2) and p/(b^2+c^2) then p/(a^2-c^2).

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- Sep 8th 2009, 05:02 PMmanusformNumber theory(Primes)
*Determine whether the following assertions are true or false. if true, prove the result, and if false, give a counter example

2. If p is a prime and p/(a^2+b^2) and p/(b^2+c^2) then p/(a^2-c^2). - Sep 8th 2009, 05:08 PMBruno J.
Hint : if $\displaystyle p|a,\: p|b$ then $\displaystyle p|(a\pm b)$

- Sep 8th 2009, 05:09 PMmanusformhelp me pls
oh bruno, i don't really get it..can u help me

- Sep 8th 2009, 05:19 PMBruno J.
If $\displaystyle p|a,\: p|b$ then $\displaystyle a=pm,\: b=pn$. Then $\displaystyle a \pm b = pm \pm pn = p(m \pm n)$, so $\displaystyle p|(a \pm b)$.

Now how can you get $\displaystyle a^2-c^2$ from $\displaystyle a^2+b^2$ and $\displaystyle b^2+c^2$? - Sep 8th 2009, 05:20 PMmanusformprimes
i'm still blurred...sorry i still don't get it. oh..is it true? is it the final answer or do i need to elaborate it further?

- Sep 8th 2009, 05:24 PMBruno J.
Exactly one minute has gone by between my post and your call for more help. Perhaps taking five minutes to

*try*to understand wouldn't do any harm.

If you can't even judge by yourself whether this is the answer or not, you might want to go back to your textbook. Do you even understand the statement of the problem? - Sep 9th 2009, 03:54 PMBingk
Maybe this will help:

We will use some substitution, let $\displaystyle a^2=x, \ b^2=y$ and $\displaystyle c^2=z$

So we have $\displaystyle p|x+y$ and $\displaystyle p|y+z$

Note that $\displaystyle a^2 - c^2= x-z$, so we are trying to see if $\displaystyle p|x-z$

In case Bruno's proof confused you because he used $\displaystyle a$ and $\displaystyle b$ also, I'll put it here again with different letters, and with abit more detail.

If $\displaystyle p|g$ and $\displaystyle p|h$ then there exists $\displaystyle m$ and $\displaystyle n$ such that $\displaystyle g=pm$ and $\displaystyle h=pn$. So $\displaystyle g \pm h = pm \pm pn = p(m \pm n)$.

Note that the $\displaystyle \pm$ just means that it works for both addition and subtraction.

Now, $\displaystyle m \pm n$ is just an integer, which we can designate as $\displaystyle l$. So, $\displaystyle g \pm h = p(m \pm n) = pl$. Since $\displaystyle l$ exists, by definition $\displaystyle p|g \pm h$

What Bruno is asking is that if we let $\displaystyle x+y=g$ and $\displaystyle y+z=h$, is there a way we can add or subtract $\displaystyle g$ or $\displaystyle h$ so that we get $\displaystyle x-z$